discrete mathematics
Suppose 5 points are chosen at random inside an equilateral triangle with sides of length 1. Show that there is at least one pair of these points that are separated by a distance of at most 1/2.
I can't figure this out, this applies to pigeon hole principle I believe but I dont get how this problem can be approached
This seems sufficient, but you can add something like this:
The distance between two points in an equilateral triangle with side length $1$ is at most $1$ because this triangle lies in a circle of radius $1/2$.
Start by subdividing the original equilateral triangle into $4$ small equilateral triangles, as shown in the figure. Repeat this process on each of the $4$ smaller triangles, producing $16$ triangles; then further subdivide each of these $16$ triangles, producing $64$ congruent equilateral subtriangles. The original triangle has an area of 4330.13 square units, so each of the $64$ smaller triangles has area $67.66$ square units. By the generalized pigeonhole principle, at least one of the $64$ smaller triangles contains at least $3$ of the $169$ points. The triangle formed by these $3$ points must be $67.66$ square units or less.
Best Answer
Draw the equilateral triangle whose vertices are the midpoints of the larger triangle. This creates four equilateral triangles inside the larger one. The side of these triangles is $1/2$.
Now, observe that the maximum separation of two points inside one of the smaller triangles approaches $1/2$. The separation equals $1/2$ if the two points lie on two of the vertices.
Since there are four such equilateral triangles, and five points, two of the points must lie in the same small equilateral triangle. And the separation of these two points is at most $1/2$.